| Exam Board | OCR |
|---|---|
| Module | FP3 (Further Pure Mathematics 3) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Groups |
| Type | Isomorphism between groups |
| Difficulty | Challenging +1.3 This is a structured group theory question requiring systematic verification of group properties, finding inverses (routine calculation), proving non-cyclicity (checking orders of elements), and identifying isomorphic subgroups. While it involves abstract algebra concepts from Further Maths, the question guides students through each step methodically with no novel insights required—just careful application of definitions and standard techniques. |
| Spec | 8.03c Group definition: recall and use, show structure is/isn't a group8.03f Subgroups: definition and tests for proper subgroups8.03g Cyclic groups: meaning of the term8.03l Isomorphism: determine using informal methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Elements row: \((1), 3, 7, 9, 11, 13, 17, 19\) | B1 | 2 or more; Ignore 1 |
| Inverse row: \((1), 7, 3, 9, 11, 17, 13, 19\) | B1 | 4 or more |
| All 7 correct | B1 | all 7 correct |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| 1 has order 1; 9, 11, 19 have order 2 | M1 | Correctly identifies order of all elements; Allow one error |
| \(3^2 = 9 \Rightarrow 3^4 = 1\) so order 4; similarly 7, 13, 17 order 4 | B1 | Justifies order for at least 1 element of order 4; must show workings towards \(a^4\) for demonstration that these elements are order 4 |
| No element of order 8 so not cyclic | A1 | www; condone "no generator" in place of "no element of order 8" |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| M1 | For two sets which both contain "1" and all (4) elements' inverses | |
| One subgroup of order 4 | B1 | |
| \(\{1, 13, 9, 17\}\) and \(\{1, 3, 9, 7\}\) | A1 | |
| M1 | For correspondence of "their" elements of same order | |
| \(1 \leftrightarrow 1, 9 \leftrightarrow 9, 3 \leftrightarrow 13, 7 \leftrightarrow 17\) | A1 | or \(3 \leftrightarrow 17, 7 \leftrightarrow 13\) |
| [5] |
# Question 4:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Elements row: $(1), 3, 7, 9, 11, 13, 17, 19$ | B1 | 2 or more; Ignore 1 |
| Inverse row: $(1), 7, 3, 9, 11, 17, 13, 19$ | B1 | 4 or more |
| All 7 correct | B1 | all 7 correct |
| **[3]** | | |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| 1 has order 1; 9, 11, 19 have order 2 | M1 | Correctly identifies order of all elements; Allow one error |
| $3^2 = 9 \Rightarrow 3^4 = 1$ so order 4; similarly 7, 13, 17 order 4 | B1 | Justifies order for at least 1 element of order 4; must show workings towards $a^4$ for demonstration that these elements are order 4 |
| No element of order 8 so not cyclic | A1 | www; condone "no generator" in place of "no element of order 8" |
| **[3]** | | |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| | M1 | For two sets which both contain "1" and all (4) elements' inverses |
| One subgroup of order 4 | B1 | |
| $\{1, 13, 9, 17\}$ and $\{1, 3, 9, 7\}$ | A1 | |
| | M1 | For correspondence of "their" elements of same order |
| $1 \leftrightarrow 1, 9 \leftrightarrow 9, 3 \leftrightarrow 13, 7 \leftrightarrow 17$ | A1 | or $3 \leftrightarrow 17, 7 \leftrightarrow 13$ |
| **[5]** | | |
---4 The group $G$ consists of the set $\{ 1,3,7,9,11,13,17,19 \}$ combined under multiplication modulo 20.\\
(i) Find the inverse of each element.\\
(ii) Show that $G$ is not cyclic.\\
(iii) Find two isomorphic subgroups of order 4 and state an isomorphism between them.
\hfill \mbox{\textit{OCR FP3 2014 Q4 [11]}}